
Since \(\alpha \lambda \)  is diffuse, a Poisson process with this distribution is
simple ([1]}), so that the above equals
\(\sum ^\infty _{n=0}&\iint f(\delta _{x_1}+\cdots +\delta _{x_n})I\lbrace x_1<\cdots <x_n\rbrace \prod ^n_{i=1}\alpha (x_i)^{-1}p(x_i,\delta _{x_1}+\cdots +\delta _{x_n})\\&\times \,\prod _{x\in \mu -(\delta _{x_1}+\cdots +\delta _{x_n})}\big (1-\alpha (x_i)^{-1}p(x,\delta _{x_1}+\cdots +\delta _{x_n})\big )\,\mu ^{(n)}(\mathrm {d}(x_1,\ldots ,x_n))\,\Pi _{\alpha \lambda }(\mathrm {d}\mu ),\) 
